# Natural Numbers form Inductive Set

## Theorem

Let $\N$ denote the natural numbers as subset of the real numbers $\R$.

Then $\N$ is an inductive set.

## Proof 1

By definition of the natural numbers:

- $\N = \displaystyle \bigcap \mathcal I$

where $\mathcal I$ is the collection of all inductive sets.

Suppose that $n \in \N$.

Then by definition of intersection:

- $\forall I \in \mathcal I: n \in I$

Since all these $I$ are inductive:

- $\forall I \in \mathcal I: n + 1 \in I$

Again by definition of intersection:

- $n + 1 \in \N$

$\blacksquare$

## Proof 2

By the given definition of the natural numbers:

- $\N = \displaystyle \bigcap \mathcal I$

where $\mathcal I$ is the collection of all inductive sets.

The result is a direct application of Intersection of Inductive Sets is Inductive Set.

$\blacksquare$

## Sources

- 2000: James R. Munkres:
*Topology*(2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers